Question

The Human Resource Manager of a large insurance company wishes to evaluate the leadership qualities of 3 manager groups- supervisors, mid-level managers and upper-level managers.

12 people from each management level were sampled.

Is there a difference on the average of the leadership scores for the three groups? Part of the leadership scores (higher is better) and ANOVA table are as follows:

Supervisor	Mid- Manager	Upper-Manager
33	29	32
43	36	39
34	32	40
29	36	46
------	-----	------
------	-----	----
-------	----	------

ANOVA TABLE  

Sources of Variation	SS	df	MS	F	p-value	F crit
Between Groups	396	2	????	3.808	---------	--------
Within Groups	1716	33	????
Total	2112	35

. Assume that the calculated value of the test statistic is 3. What is the conclusion for the ANOVA test? Use the critical F-value from Question 11.

(Question 11:

The critical value of the ANOVA test at 1% alpha is:

5.31)

Answer 1

Given ANOVA table:

#Formula:

i) MS Between Groups = (SS Between Groups/ df Between Groups)

ii) MS Within Groups = (SS Within Groups/ df Within Groups)

iii) F-stat = (MS Between Groups/MS Within Groups) ]

Calculation:

i) MS Between Groups = (396/2) = 198

ii) MS Within Groups = (1716/33) = 52

iii) F-stat = (198/52) = 3.807692308 = 3.808

Now we find the F-critical value at level and DF = (2, 33)

[ Value are getting from F critical value table with corresponding and (d.f) = (2, 33) ]

We find p-value by Excell

=FDIST(5.31,2,33)

So we get p-value = 0.0325 [round to three decimal places]

So now complete ANOVA table is given by,

Sources of variation	SS	df	MS	F	p-value	F crit
Between Groups	396	2	198	3.807692308	0.032515	5.312029
Within Groups	1716	33	52
Total	2112	35

the critical value of the ANOVA test at 1% alpha is: 5.31

here P-value >0.01 hence we fail to reject the null hypothesis

Conclusion : At α = 0.01, there is not sufficient evidence to conclude that there is significant difference in the average leadership scores for the three groups.